1. **State the problem:** We have 7 red counters and some blue counters in a bag. The total number of counters is unknown. We know there are more blue counters than red counters. 2. **Given:** - Number of red counters $= 7$ - Number of blue counters $= b$ (unknown) - Total counters $= 7 + b$ - Probability of drawing one red and one blue counter in two draws $= \frac{7}{15}$ - Blue counters are more than red counters, so $b > 7$ 3. **Formula for probability of one red and one blue in two draws:** The event "one red and one blue" can happen in two ways: - First draw red, second draw blue - First draw blue, second draw red Probability = $$\frac{7}{7+b} \times \frac{b}{7+b-1} + \frac{b}{7+b} \times \frac{7}{7+b-1}$$ 4. **Simplify the probability expression:** $$= \frac{7b}{(7+b)(6+b)} + \frac{7b}{(7+b)(6+b)} = \frac{14b}{(7+b)(6+b)}$$ 5. **Set the probability equal to given value and solve for $b$:** $$\frac{14b}{(7+b)(6+b)} = \frac{7}{15}$$ Multiply both sides by $(7+b)(6+b)$: $$14b = \frac{7}{15} (7+b)(6+b)$$ Multiply both sides by 15: $$210b = 7 (7+b)(6+b)$$ Divide both sides by 7: $$30b = (7+b)(6+b)$$ Expand right side: $$30b = 42 + 7b + 6b + b^2 = 42 + 13b + b^2$$ Bring all terms to one side: $$0 = 42 + 13b + b^2 - 30b$$ $$0 = b^2 - 17b + 42$$ 6. **Solve quadratic equation:** $$b^2 - 17b + 42 = 0$$ Use quadratic formula: $$b = \frac{17 \pm \sqrt{17^2 - 4 \times 1 \times 42}}{2} = \frac{17 \pm \sqrt{289 - 168}}{2} = \frac{17 \pm \sqrt{121}}{2} = \frac{17 \pm 11}{2}$$ Two solutions: - $$b = \frac{17 + 11}{2} = \frac{28}{2} = 14$$ - $$b = \frac{17 - 11}{2} = \frac{6}{2} = 3$$ 7. **Check condition $b > 7$:** Only $b = 14$ satisfies this. 8. **Find total counters:** $$7 + 14 = 21$$ **Final answer:** The total number of counters in the bag is 21.